Jacaranda (Engineering) 3333
Mail Code
Phone: 818.677.6448
E

mail:
lcaretto@csun.edu
8348
Fax: 818.677.7062
College of Engineering and Computer Science
Mechanical Engineering Department
Mechanical Engineering 37
5
Heat Transfer
Spring 2007
Number
1
7629
Instructor: Larry Caretto
Solutions to
Exercise
Nine
–
Internal
Forced
Convection
and
Free
Convection
1.
Hot air at atmospheric pressure and
85°C
enters a 10

m

long uninsulated square duct of
cross section 0.15 m by 0.15 m that passes
through the attic of a house at a rate of 0.10
m
3
/s.
The duct is observed to be nearly
isothermal at 70°C.
Determine the exit
temperature of the air and the rat
e of heat loss
from the duct to the air space in the attic.
(Problem and figure P
8

45
from Çengel, Heat
and Mass Transfer.)
We will assume a mean temperature of 80
o
C for the flow to evaluate the properties. We will
check this assumption after we compute
the exit temperature. The properties of air are found
from Table A

15:
= 0.9994 kg/m
3
,
k = 0.02593 W/m
∙
o
C,
= 1.774x10

5
m
2
/s,
c
p
= 1008 J/kg∙
o
C,
and Pr = 0.7154.
The square duct with a side a = 0.15 m has a hydraulic diameter found as follows.
m
a
a
a
p
A
D
h
15
.
0
4
4
4
2
Once the velocity is found from the volume flow rate we can compute the Reynolds number.
s
m
m
s
m
a
V
A
V
V
c
444
.
4
15
.
0
10
.
0
2
3
2
4
2
5
10
179
.
3
10
774
.
1
15
.
0
444
.
4
Re
x
s
m
x
m
s
m
VD
h
h
For
this Reynolds number the flow is turbulent so the entry length will be about 10 diameters or
1.5 m
. This is only 15% of the total duct length so we can assume fully developed turbulent flow.
We will use the Dittus

Boelter equation (8

68) with a Prandtl number exponent n = 0.3 because
the cooler duct temperature is cooling the fluid, to find the Nusse
lt number and the heat transfer
coefficient.
16
.
83
7154
.
0
10
179
.
3
023
.
0
Pr
Re
023
.
0
3
.
0
8
.
0
4
3
.
0
8
.
0
x
Nu
C
m
W
C
m
W
m
D
kNu
h
o
o
h
2
37
.
16
02593
.
0
15
.
0
16
.
83
Exercise
nine
solutions
ME 3
75
, L. S. Caretto, Spring 200
7
Page
2
We can compute the exit temperature of the air by the formula usual formula with NTU. We first
compute
the mass flow rate and
the surface area for heat transfer (
the four sides of the duct)
.
s
kg
s
m
m
kg
V
m
09994
.
0
10
.
0
9994
.
0
3
3
2
6
10
15
.
0
4
4
m
m
m
aL
A
s
9750
.
0
1
1008
09994
.
0
6
37
.
16
2
2
J
s
W
s
kg
J
s
kg
m
C
m
W
c
m
hA
NTU
o
p
s
9750
.
0
85
70
70
e
C
C
C
e
T
T
T
T
o
o
o
NTU
in
s
s
out
75.7
o
C
The mean temperature between the inlet and the outlet is (85
o
C + 75.7
o
C)/2 = 80.35
o
C, which is
close enough to our assumed me
an temperature of 80
o
C so that we do not have to repeat the
calculation
.
We can now find the heat transfer from the
first law equation
.
C
C
J
s
W
s
kg
J
s
kg
T
T
c
m
Q
o
o
in
out
p
85
7
.
75
1
1008
09994
.
0
=
–
941
W
9

36
E
Consider an industrial furnace that resembles
a 13

ft

long horizontal cylindrica
l enclosure 8
ft in diameter whose end surfaces are well
insulated. The furnace burns natural gas at a
rate of 48 therms/h.
The combustion
efficiency of the furnace is 82 percent (i.e., 18
percent of the chemical energy of the fuel is
lost through the flu
e gases as a result of
incomplete combustion and the flue gases
leaving the furnace at high temperature).
If
the heat loss from the outer surfaces of the
furnace by natural convection and radiation
is not to exceed 1 percent of the heat
generated inside,
determine the highest
allowable surface temperature of the furnace.
Assume the air and wall surface
temperature of the room to be 75
o
F, and take the emissivity of the outer surface of the
furnace to be 0.85.
If the cost of natural gas is $1.15/therm and
the furnace operates 2800
h per year, determine the annual cost of this heat loss to the plant.
(Problem and Figure
P9

36
E from Çengel,
Heat and Mass Transfer
.)
We have an iterative solution because we do not know the surface temperature. If we assume
th
at the surface temperature is 140
o
F we can find the properties for air at the mean temperature
of
(140
o
F +
7
5
o
F
)/2 =
107
.5
o
F
from Table A

15E:
k = 0.01
546
Btu/h
ft
o
F,
= 0.1
852
x10

3
ft
2
/s,
and Pr = 0.7
249
. At the mean temperature of
107
.5
o
R
= 5
67
.17 R, the expansion coefficient,
=
1/T = 0.001
763
R

1
. Using the temperature difference of
T =
140
o
F
–
75
o
F =
6
5
o
F =
6
5 R and
the characteristic length as
the diameter D
=
8
ft, we can find the Rayleigh number.
10
2
2
3
3
2
2
3
10
991
.
3
7249
.
0
10
1852
.
0
8
65
001763
.
0
174
.
32
Pr
x
ft
x
s
ft
R
R
s
ft
TD
g
Ra
D
Exercise
nine
solutions
ME 3
75
, L. S. Caretto, Spring 200
7
Page
3
This value of Ra
L
is in the correct range to use equation 9

25 in Çengel for the Nusselt number from
which we can find the heat transfer coefficient.
8
.
376
7249
.
0
559
.
0
1
10
991
.
3
387
.
0
6
.
0
Pr
559
.
0
1
387
.
0
6
.
0
2
27
/
8
16
/
9
6
/
1
10
2
27
/
8
16
/
9
6
/
1
x
Ra
Nu
D
F
ft
h
Btu
F
ft
h
Btu
ft
L
kNu
h
o
o
2
7287
.
0
01546
.
0
8
8
.
376
The surface area for heat transfer is the area of the cylindrical surface (assuming that the well

insulated ends do not make a significant contribution to the heat transfer.
)
2
7
.
326
13
8
ft
ft
ft
DL
A
s
The heat transfer through the walls will be 1% of
the heat generated in the furnace. Since the
furnace is 82% efficient and has a heat input of 48 therms/h, the heat generation within the
furnace is (48 therms/h)(10
5
Btu/therm)(82%) = 3.936x10
6
Btu/h. Only 1% of this heat or
3
.
93
6
x10
4
Btu/h will leave t
hrough the furnace walls by convection and radiation. The surface
temperature can be found from the following equation for the combined modes of heat transfer.
4
4
4
2
8
2
2
4
4
2
4
67
.
534
10
1717
.
0
85
.
0
67
.
534
7287
.
0
5
.
120
7
.
326
10
936
.
3
R
T
R
ft
h
Btu
x
R
T
R
ft
h
Btu
ft
h
Btu
T
T
T
T
h
ft
h
Btu
x
A
Q
q
s
s
s
s
s
Using an iterative solution procedure such as the goal seek method of E
xcel gives the solution to
this equation as T
s
= 601.8 R = 141.8
o
R. We could repeat the entire calculation with this value of
T
s
to compute the properties and the Rayleigh number, instead of the value of 140
O
F that we
assumed. However, the differences wi
ll be negligible, and we conclude that the surface
temperature the furnace is
T
s
=
141.8
o
C
.
Over a 2800 hour per year operating
period there would be a heat loss of
(
3,935x10
4
Btu/h)(2800 h/yr) =
1.102x10
8
Btu/yr = 1.102x10
3
therms/yr. For natural gas th
at costs
$1.15/therm, this heat loss is an
annual
cost
of
$1267/yr
.
9

58E
The figure at the right shows a
6

in

wide and 8

in

high vertical hot
surface in 78
o
F air
that
is to be
cooled by a heat sink with equally
spaced fins of rectangular profile.
The fins
are 0.08 in thick and 8 in
long in the vertical direction and
Exercise
nine
solutions
ME 3
75
, L. S. Caretto, Spring 200
7
Page
4
have a height of 1.2 in from the base.
Determine the optimum fin spacing and the rate of
heat transfer by natural convection from the heat sink if the base temperature is 180
o
F.
(Problem
and
figure from
Çengel,
Heat and Mass Transfer
.)
W
e can find the properties for air at the mean temperature of
(1
8
0
o
F + 7
8
o
F)/2 =
129
o
F
from
Table A

15E:
k = 0.015
97
Btu/h
ft
o
F,
= 0.1
975
x10

3
ft
2
/s, and Pr = 0.72
17
. At the mean
temperature of
129
o
F
=
588
.
6
7 R, the expansion coefficient,
= 1/T = 0.001
699
R

1
. Using the
temperature difference of
T = 1
8
0
o
F
–
7
8
o
F =
102
o
F =
102
R and the characteristic length
i
s the
fin height, L
= 8
in = 0.6667 ft
, we can find the Rayleigh number.
7
2
2
3
3
2
2
3
10
058
.
3
7217
.
0
10
1975
.
0
6667
.
0
102
001699
.
0
174
.
32
Pr
x
ft
x
s
ft
R
R
s
ft
TD
g
Ra
T
h
e optimum fin spacing, S
opt
, is given by equation (9

32) in Çengel.
ft
x
ft
Ra
L
S
opt
02433
.
0
10
058
.
3
6667
.
0
714
.
2
714
.
2
7
4
1
=
0.292
in
When the optimum fin spacing is used, equation (9

33) in Çengel tells us that the Nusselt number based
on the fin spacing is a constant equal to 1.307.
This gives the heat transfer coefficient.
F
ft
h
Btu
F
ft
h
Btu
ft
S
kNu
h
o
o
opt
2
7287
.
0
01597
.
0
02433
.
0
307
.
1
The
number of fins, n, is found from the fin thickness and the fin spacing compared to the total
width.
fins
fin
in
fin
in
in
n
16
08
.
0
292
.
0
6
Each of these fins has two side areas of 1.2 in by 18 in plus
an outer edge that is 0.
08 in thick
and has a total length on three sides of 1.2 in + 1.2 in + 8 in = 10.4 in. This gives the total fin
area.
2
2
226
.
2
5
.
320
08
.
0
4
.
10
2
.
1
8
2
16
ft
in
in
in
in
in
A
s
The
convective heat transfer is given by the usual equation
F
F
ft
F
ft
h
Btu
T
T
hA
Q
o
o
o
s
s
78
180
227
.
2
8578
.
0
2
2
196
Bt
u/h
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